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Mathematics Portfolio Sl Essays Mathematics Portfolio Sl Essay Mathematics Portfolio Sl Essay Mathematics Standard Level Teacher: Mr. Lazaro Name: Fatema Ismailjee IB 1 2011 Sequence is a set of things (usually numbers) that are in order. e. g. 1, 2, 3, 4, Where 1 is the first term, 2 is the second term and so on. ( in the end means that the sequence goes on forever. Three dots in the middle e. g. 1, 2, 3 7, 8 indicate that the pattern continues until the next number appears. There is finite and infinite sequence, infinite sequence is when the sequence has no end and finite is a set with a function e. g. {1, 3, n} Calculating specific terms leads to an nth  term formula. Before creating a rule of calculation, you need to realize that sequences are functions with the specific domain of the counting numbers {1, 2, 3, 4, 5, }. So the n replaces x  as the input variable and instead of writing  y, we use  an  as the output variable.Arithmetic sequence: the difference between one term and the next is a constant in arithmetic sequence. The general formula is an  = a1à ‚  + (n 1) d Geometric sequence: A geometric sequence is a group of numbers where each term after the first is found by multiplying the previous one by a fixed non zero number called common ratio. The general formula is an = a1 ? rn-1 Series is the sum of terms of a sequence. Sn = x1 + x2 +. xn Arithmetic series: The general formula is Sn  = n/2(a1  + an) Geometric series: a series which has a constant ratio between terms.The general formula is Sn = a1 (1 – rn) 1 r TRIANGULAR NUMBERS Triangular number is the number of dots in an equilateral triangle uniformly filled with dots. This is an investigation task whereby I will try to find number of shapes of geometric figures which form triangular numbers. I will use different sources of information to attain shapes and figures. For the calculations required, different math techniques will be used for the different shape obtained. Aim In this task I will consider geometric shapes which lead to special numbers.The simplest exa mples of these are square numbers, 1, 4, 9, 16, which can be represented by squares of side 1, 2, 3 and 4. The following diagrams show a triangular pattern of evenly spaced dots. The numbers of dots in each diagram are examples of triangular numbers (1, 3, 6, ). .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 3 6 10 15 There is a sequence of the number of dots in the triangular shape above.Complete the triangular sequence with three more terms. . . . . . . . . . . . . . . . . . . . . . 21 dots . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 dots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 dots Find a general statement that represents the nth triangular number in terms of n. In words: The top row has one dot and each successive row under it has one more dot.Using the formula: 1. Find the common difference between the numbers in the sequence. 2. Use the general formula tn = an2 + bn + c. 3. Three equations will be forme d. Using the elimination method find the coefficients i. e. a, b and c. 4. Substitute in the general formula. The general statement can be reached by following the steps above. Common difference: d= U2 – U1 = U3 – U2 1, 3, 6, 10, 15, d= 3-1 = 2 6-3 = 3 10-6 = 4 15-10 = 5 d= 3-2 = 1 4-3 = 1 5-4 = 1 The difference in terms is found in the second stage so the formula will be n2 . 2 Testing: n = 1 , triangular number = 1 22 = 12 1 12 = 12 n = 2, triangular number = 3 n22 = 92 3 92 = 32 32 12 = 12 12 so this will be 12 n therefore, 2 12 n2 = 12n As the common difference in the second stage is 1, it can be deduced that the formula for the nth term is a quadratic equation. I will use the general formula to find the nth term, tn = an2 + bn + c where a and b are the coefficients and c is constant and n is the number of term. 2 n2 = 12 n = n2 + n 2 When n = 1 1 = a (1)2 + b (1) + c 1 = a + b + c . (i) n = 2 3 = a (2)2 + b (2) + c 3 = 4a + 2b + c . (ii) n = 3 6 = a (3)2 + b (3 ) + c 6 = 9a + 3b + c . (iii) Using the elimination method: 3 = 4a + 2b + c 6 = 9a + 3b + c 1 = a + b + c 3 = 4a + 2b + c 2 = 3a + b 3 = 5a + b Now that two equations are obtained: To find the variables i. e. a, b one of them is eliminated.In this case the equations are being subtracted. b will be eliminated first in order to find a. Substitute the values of a in the equation to find the value of b. 3 = 5a + b 3 = 5a + b 1 = a + b + c 2 = 3a + b 3 = 5(1 ) + b 1 = 1 + 1 +c 2 2 2 1 = 2a 3 5 = b 2 2 2 a = 1 b = 1 c = 0 2 Therefore the formula for finding the nth term will be as follows: tn = 1n2 + 1n 2 2 tn = n2 + n 2 Use of technology to find the general statement: Calculator used: CASIO fx-9750 GA PLUS n| 1| 2| 3| 4| 5| 6| 7| y| 1| 3| 6| 10| 15| 21| 28| Let n = x 1. Select STAT. 2. Encode values for x in list 1 and for y in list 2. 3. Select GRPH (by pressing F1). 4. Select GPH1 (by pressing F1 again). 5. Select x^2 (by pressing F3). The display will show: a = 1 2 b = 1 c = 0 y = ax 2 + bx + c 1n2 + 1n y = 2 2 = n2 + n 2 Consider stellar (star) shapes with p vertices, leading to p-stellar numbers. The first four representations for a star with six vertices are shown in the four stages S1 – S4 below. The 6-stellar number at each stage is the total number of dots in the diagram. Find the number of dots (i. e. the stellar number) in each stage up to S6. Stellar numbers are figurate number, based on the number of dots of units that can fit in a centred hexagon or star shapes. S1 – S4 are the numbers of dots in the stars.To find up to S6 find the common difference (d) followed by the addition of numbers of star in the previous star. S1 has 1 dot S2 has 13 dots S3 has 37 dots S4 has 73 dots Find the common difference between the terms. d = S2 – S1 S3 – S2 d = 13 – 1 = 12 37 – 13 = 24 73 – 37 = 36 As the difference is not constant, find the difference within the answers. d = 36 – 24 = 12 13 – 12 = 12 The c ommon difference is 12. S 5 = 36 + 12 = 48 73 + 48 = 121dots S6 = 48 + 12 = 60 121 + 60 = 181dots Find an expression for the 6-stellar number at stage S7. As shown above, the common difference is 12.As it’s a sequence it follows the same trend therefore: To find the next number of dots in the sequence, add it with 12 first and from the second star add it with the multiples of 12, i. e. 24, 36, 48 etc. S6 = 48 + 12 = 60 = S5 + 60 = 121 + 60 = 181 S7 = 60 + 12 = 72 = S6 + 72 = 181 + 72 = 253 S7 = 253 Find a general statement for the 6-stellar number at stage Sn in terms of n. Use the same general formula to obtain the three equations: The general formula: tn = an2 + bn + c When n = 1 1 = a (1)2 + b (1) + c 1 = a + b + c . (i) n = 2 13 = a (2)2 + b (2) + c 13 = 4a + 2b + c . (ii) = 3 37 = a (3)2 + b (3) + c 37 = 9a + 3b + c . (iii) Using the elimination method: 37 = 9a + 3b + c 13 = 4a + 2b + c 13 = 4a + 2b + c 1 = a + b + c 24 = 5a + b 12 = 3a + b After attaining two equations, either of the coefficients should be eliminated. b in this case which will lead us to find a. Substitute value of a in the equation to find b. Hence, substitute values of a and b for c. 24 = 5a + b 24 = 5a + b 1 = a + b + c 2 = 3a + b 24 = 5(6) + b 1 = 6 + (-6) + c 12= 2a 24 – 30 = b 1 – 0 = c 2 2 a = 6 b = -6 c = 1 Substitute a, b and c in the general statement. General statement: tn = 6n2 – 6n + 1 Now repeat the steps above for other values of p Considering stellar (star) shapes when p=7 and when p=8 leading to p-stellar numbers. p = 7 Find the number of dots (i. e. the stellar number) in each stage up to S6. S1 has 1 dot S2 has 15 dots S3 has 43 dots S4 has 85 dots d = 15 – 1 = 14 3 – 15 = 28 85 – 43 = 42 d = 42 – 28 = 14 28 – 14 = 14 e. g. S4 = 43 + 14 = 42 S3 + 42 43 + 42 = 85 dots S 5 = 42 +14 =56 S4 + 56 85 + 56 = 141dots S6 = 56 +14 = 70 S5 + 70 141 + 70 = 211dots Find an expression for the 6-stellar number at stage S7 . As shown above, the common difference is 14. As it’s a sequence it follows the same trend therefore: To find the next number of dots in the sequence, add it with 2 first and from the second star add it with the multiples of 2, i. e. 14, 28, 42 etc. S7 = 70 + 14 = 84 S6 + 84 211 + 84 = 2955dots S7 = 295dots Find a general statement for the 6-stellar number at stage Sn in terms of n.To find the three equations, use the general formula tn = an2 + bn + c. When n = 1 1 = a (1)2 + b (1) + c 1 = a + b + c . (i) n = 2 15 = a (2)2 + b (2) + c 15 = 4a + 2b + c . (ii) n = 3 43 = a (3)2 + b (3) + c 43 = 9a + 3b + c . (iii) Three equations are obtained, to find a, b and c, the equations need to be solved. Elimination method is one of the ways from which we can attain the coefficients and constant. Using elimination method: Firstly, we need to remain with two equations at the end so subtract equations (equation iii – ii and equation ii – i) and two will be remained. 3 = 9a + 3b + c 15 = 4a + 2b + c 15 = 4a + 2b + c 1 = a + b + c 28 = 5a + b 14 = 3a + b Now that there are two equations, find a and b. Subtract the equation to eliminate one variable. After one is found, the other can be easily found by substituting the value of variable attained in the equation. 28 = 5a + b 28 = 5a + b 1 = a + b + c 14 = 3a + b 28 = 5(7) + b 1 = 7 + (-7) + c 4= 2a 28 – 35 = b 1 – 0 = c 2 2 a = 7 b = -7 c = 1 Substitute a, b and c in the general statement. General statement: tn = 7n2 – 7n + 1 p = 8 S1 has 1 dot S2 has 17 dots S3 has 49 dots S4 has 97 dots Find the common difference: d = 17 – 1 = 16 49 – 17 = 32 97 – 49 = 48 As the difference is not constant, subtract the answers to find the common difference. d = 32 – 16 = 16 48 – 32 = 16 To find the following number in the star e. g. S4 = 32 + 16 = 48 S3 + 48 49 + 48 = 97 dots The common difference is 16.If observed carefully the number is found by adding it with mu ltiples of 16 i. e. 32, 48, 64, 80 etc. S 5 = 48 +16 = 64 S4 + 64 97 + 64 = 161dots S6 = 64 + 16 = 80 S5 + 80 161 + 80 = 241dots Find an expression for the 6-stellar number at stage S7. As shown above, the common difference is 16. As it’s a sequence it follows the same trend therefore: To find the next number of dots in the sequence, add it with 16 first and from the second star add it with the multiples of 3, i. e. 32, 48, 64 etc. S7 = 80 + 16 = 96 S6 + 96 241 + 96 = 337dots S7 = 337dots Find a general statement for the 6-stellar number at stage Sn in terms of n.I will use the same general formula to obtain the three equations: The general formula: tn = an2 + bn + c When n = 1 1 = a (1)2 + b (1) + c 1 = a + b + c . (i) n = 2 17 = a (2)2 + b (2) + c 17 = 4a + 2b + c . (ii) n = 3 49 = a (3)2 + b (3) + c 49 = 9a + 3b + c . (iii) Using the elimination method: 49 = 9a + 3b + c 17 = 4a + 2b + c 17 = 4a + 2b + c 1 = a + b + c 32 = 5a + b 16 = 3a + bNow there are two equations, so b has to eliminated by subtracting the two equations to find a. Once, a is obtained one of the equation has to be chosen and substitute the value of a in it. Hence b is obtained. 32 = 5a + b 32 = 5a + b 1 = a + b + c 16 = 3a + b 32 = 5(8) + b 1 = 3 + (-3) + c 16 = 2a 32 – 40 = b 2 2 a = 8 b = -8 c = 1 Substitute a, b and c in the general statement. General statement: tn = 8n2 – 8n + 1Hence, produce the general statement, in terms of p and n,that generates the sequence of p-stellar numbers for any value of p at stage Sn. The general statements produced are: tn = 6n2 – 6n + 1 tn = 7n2 – 7n + 1 tn = 8n2 – 8n + 1 I observed it, and reached to the conclusion that for all the three statements the number of p and number of coefficient that is a and b is the same. Therefore the general statement in terms of p and n that generates the sequence of p-stellar numbers for any value of p at stage Sn is: tn = pn2 – pn + 1 Test the validity of the general st atement. When p = 5 S1 has 1 dot S2 has 11 dots S3 has 31 dotsS4 has 61 dots Common difference: d = 11 – 1 = 10 31 – 11 = 20 61 – 31 = 30 As the difference is not constant, subtract it within the answer obtained: d = 30 – 20 = 10 20 – 10 = 10 As seen, the common difference is 5. As it’s a sequence it follows the same trend therefore: To find the next number in the sequence, add it with multiples of 5 i. e. 10, 15, 20, etc. S 5 = 30 + 10 = 40 61 + 40 = 101dots S6 = 40 + 10 = 50 = S5 + 50 = 101 + 50 = 151dots S7 = 50 + 10 = 60 = S6 + 60 = 151 + 60 = 211 S7 = 211dots The general formula: tn = an2 + bn + c When n = 1 1 = a (1)2 + b (1) + c 1 = a + b + c . (i) n = 2 1 = a (2)2 + b (2) + c 11 = 4a + 2b + c . (ii) n = 3 31 = a (3)2 + b (3) + c 31 = 9a + 3b + c . (iii) Using the elimination method: 31 = 9a + 3b + c 11 = 4a + 2b + c 11 = 4a + 2b + c 1 = a + b + c 20 = 5a + b 10 = 3a + b 20 = 5a + b 20 = 5a + b 1 = a + b + c 10 = 3a + b 20 = 5(5) + b 1 = 6 + (-6) + c 10= 2a 20 – 25 = b 1 – 0 = c 2 a = 5 b = -5 c = 1 Substitute a, b and c in the general statement. General statement: tn = 5n2 – 5n + 1 Below are the values of the formula Sn= 5n2 5n +1 n = Stage number in the 5-stellar shape. | y= total number of dots at stage ‘n’. | 1| 1| 2| 11| 3| 31| 4| 61| 5| 101| 6| 151| 7| 211| If we want to know the number of dots in the 5th term using the formula we replace n with 5 based on the formula Sn= 5n2 – 5n + 1 Sn= 5n2 – 5n + 1 Sn= 5 (5)2 – 5 (5) +1 Sn= 101 Limitation: * The value of p should be greater than or equal to 4 i. e. p ? 4. The value of p cannot be negative. It must be a positive integer. The general statement has some limitations as listed above. It is an arithmetic series as seen. It is derived from the equations generated in the 5, 6, 7-stellar shape. The coefficients in each question are equal to the corresponding stellar number p. References: Sequences.   Math Is Fun Maths Resources. Web. 12 Mar. 2011. Help for a Generic Formula for a Stellar Pattern.? Yahoo! Answers.   Yahoo! Answers Home. Web. 12 Mar. 2011. Triangular Number ENotes. com Reference.   ENotes Literature Study Guides, Lesson Plans, and More. Web. 11 Mar. 2011.